Let $$ f(w)=\frac{w(1-i)-(i-1)}{w-1} $$, where $w$ is the left hand plane.
What is the image of this map?
The answer should be $|z|^2<2$ if I did everything before correctly. This is a show that question. I was asked to show that the image of $f(z)=\frac{z-(i-1)}{z-(1-i)}$ when $|z|^2<2$ is a left half plane. So (using $w=f(z)$ I took the inverse, and now I need to find the image of the inverse map.
Take a point on the boundary of the region $|z|^2=2$, i.e. a circle, centre at the origin and radius $\sqrt{2}$ and let this point be
$$z=\sqrt{2}(\cos\theta+i\sin\theta)$$
Then its image is $$w=\frac{\sqrt{2}\cos\theta+1+i(\sqrt{2}\sin\theta-1)}{\sqrt{2}\cos\theta-1+i(\sqrt{2}\sin\theta+1)}$$ $$=\frac{\sqrt{2}\cos\theta+1+i(\sqrt{2}\sin\theta-1)}{\sqrt{2}\cos\theta-1+i(\sqrt{2}\sin\theta+1)}\times\frac{\sqrt{2}\cos\theta-1-i(\sqrt{2}\sin\theta+1)}{\sqrt{2}\cos\theta-1-i(\sqrt{2}\sin\theta+1)}$$
It is readily seen that the real part of this expression is identically zero, whereas the imaginary part is a function of $\theta$, meaning that the image of the circle is the imaginary axis.
Now take one point inside the circle, i.e. satisfying the given inequality, for example, $z=0$.
The image of this point is $$w=\frac{1-i}{-1+i}=-1.$$ Therefore the image of the region inside the original circle is the region to the left of the imaginary axis.